Jensen Mclean

2022-09-04

Simple fraction , can someone help

$=\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}+\frac{1}{(k+1)(k+3)}$

how can i simplify above fraction please, can someone guide me from beginning to the end?

help am a beginner

i took the LCM of both to be 4(k+1)(k+2)(k+3)

4 into the LCM =(k+1)(k+2)(k+3) 2(k+1)(k+2) in to the LCM =2(k+3) (k+1)(k+3) in to the LCM = 4(k+2)

therefore $=\frac{(k+1)(k+2)(k+3)-2(k+3)(2k+3)+4(k+2)}{4(k+1)(k+2)(k+3)}$

$=\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}+\frac{1}{(k+1)(k+3)}$

how can i simplify above fraction please, can someone guide me from beginning to the end?

help am a beginner

i took the LCM of both to be 4(k+1)(k+2)(k+3)

4 into the LCM =(k+1)(k+2)(k+3) 2(k+1)(k+2) in to the LCM =2(k+3) (k+1)(k+3) in to the LCM = 4(k+2)

therefore $=\frac{(k+1)(k+2)(k+3)-2(k+3)(2k+3)+4(k+2)}{4(k+1)(k+2)(k+3)}$

dheasca8d

Beginner2022-09-05Added 6 answers

First you want to put everything over a common denominator, which should be a multiple of all the denominators. Here you would go for $4(k+1)(k+2)(k+3)$

Now you need to multiply top and bottom of each fraction in turn by the appropriate amount to make the bottom what you are aiming for. So you get

$\frac{3(k+1)(k+2)(k+3)}{4(k+1)(k+2)(k+3)}-\frac{(2k+3)2(k+3)}{2(k+1)(k+2)2(k+3)}+\frac{4(k+2)}{(k+1)(k+3)4(k+2)}$

which equals

$\frac{3(k+1)(k+2)(k+3)-2(2k+3)(k+3)+4(k+2)}{4(k+1)(k+2)(k+3)}.$

Next you need to expand out each product in the top, and collect like terms. When you've done that you should try to factorise the top and cancel any common factor with the bottom. If you've done it right, you should get a common factor of $(k+1)$

Now you need to multiply top and bottom of each fraction in turn by the appropriate amount to make the bottom what you are aiming for. So you get

$\frac{3(k+1)(k+2)(k+3)}{4(k+1)(k+2)(k+3)}-\frac{(2k+3)2(k+3)}{2(k+1)(k+2)2(k+3)}+\frac{4(k+2)}{(k+1)(k+3)4(k+2)}$

which equals

$\frac{3(k+1)(k+2)(k+3)-2(2k+3)(k+3)+4(k+2)}{4(k+1)(k+2)(k+3)}.$

Next you need to expand out each product in the top, and collect like terms. When you've done that you should try to factorise the top and cancel any common factor with the bottom. If you've done it right, you should get a common factor of $(k+1)$

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